Answer:
Step-by-step explanation:Let's solve the system of equations using a systematic approach without resorting to excessive calculation.
Given:
�
+
�
+
�
=
1000
a+b+c=1000
�
5
+
�
6
+
�
7
=
160
5
a
+
6
b
+
7
c
=160
First, from (1), we can express
�
=
1000
−
�
−
�
c=1000−a−b.
Now, substituting
�
c in equation (2), we get:
�
5
+
�
6
+
1000
−
�
−
�
7
=
160
5
a
+
6
b
+
7
1000−a−b
=160.
Multiplying everything by 210 (which is the LCM of 5, 6, and 7) to get rid of the denominators:
42
�
+
35
�
+
30
(
1000
−
�
−
�
)
=
33600
42a+35b+30(1000−a−b)=33600
Expanding and combining like terms:
42
�
+
35
�
+
30000
−
30
�
−
30
�
=
33600
42a+35b+30000−30a−30b=33600
12
�
+
5
�
=
3600
12a+5b=3600
12
�
=
3600
−
5
�
12a=3600−5b
�
=
300
−
5
�
12
a=300−
12
5b
For "a" to be an integer,
5
�
5b must be divisible by 12. The factors of 12 are 1, 2, 3, 4, 6, and 12. Only 4 and 12 can be factors of 5b since 5 is prime. Thus,
�
=
4
×
3
=
12
b=4×3=12 or
�
=
12
×
3
=
36
b=12×3=36 are the possible values.
If
�
=
12
b=12,
�
=
300
−
5
(
12
)
12
=
295
a=300−
12
5(12)
=295. From the first equation,
�
=
1000
−
295
−
12
=
693
c=1000−295−12=693.
If
�
=
36
b=36,
�
=
300
−
5
(
36
)
12
=
285
a=300−
12
5(36)
=285. From the first equation,
�
=
1000
−
285
−
36
=
679
c=1000−285−36=679.
Thus, there are two integer solutions to the system of equations:
(a, b, c) = (295, 12, 693) and (285, 36, 679).